In the previous article we learned about ripple factor in power supply circuits, here we continue and evaluate the formula for calculating ripple current, and consequently the filter capacitor value for eliminating the ripple content in the DC output.
The previous post explained how a DC content after rectification may carry the maximum possible amount of ripple voltage, and how it may be reduced significantly by using a smoothing capacitor.
Although the final ripple content which is the difference between the peak value and the minimum value of the smoothed DC, never seem to eliminate completely, and directly relies on the load current.
In other words if the load is relatively higher, the capacitor begins losing its ability to compensate or correct the ripple factor.
Standard Formula for Calculating Filter Capacitor
In the following section we will try to evaluate the formula for calculating filter capacitor in power supply circuits for ensuring minimum ripple at the output (depending on the connected load current spec).
C = I / (2 x f x Vpp)
where I = load current
f = input frequency of AC
Vpp = the minimum ripple (the peak to peak voltage after smoothing) that may be allowable or OK for the user, because practically it's never feasible to make this zero, as that would demand an unworkable, non-viable monstrous capacitor value, probably not feasible for anybody to implement.
Let's try to understand the relation between load current, ripple and the optimal capacitor value from the following evaluation.
Relation Between Load Current, Ripple, and Capacitor Value
In the mentioned formula we can see that the ripple and the capacitance are inversely proportional, meaning if the ripple needs to be minimum, the capacitor value needs to increase and vice versa.
Suppose we agree to a Vpp value that's, say 1V, to be present in the final DC content after smoothing, then the capacitor value may be calculated as shown below:
Example:
C = I / 2 x f x Vpp (assuming f = 100Hz and load current requirement as 2amp))
Vpp should be ideally always a one because expecting lower values can demand huge unpracticable capacitors values, so "1" Vpp can be taken as a reasonable value.
Solving the above Formula we get:
C = I / (2 x f x Vpp)
= 2 / (2 x 100 x 1) = 2 / 200
= 0.01 Farads or 10,000uF (1Farad = 1000000 uF)
Thus, the above formula clearly shows how the required filter capacitor may be calculated with respect to the load current and the minimum allowable ripple current in the DC component.
By referring to the above solved example, one may try varying the load current, and/or the allowable ripple current and easily evaluate the filter capacitor value accordingly for ensuring an optimal or the intended smoothing of the rectified DC in a given power supply circuit.
sir, namaskaar hamein vpp apne load ke hisaab se kitna rakhna chahiye ya phir vpp hamesha har circuit mein 1 hi rahega.kripya bataye
Ashu, Vpp can be always assumed to be 1 which is an acceptable value for all types of filter capacitor calculations.
Lets look at a circuit that has 42Vp from a fwbr. Output current is up to 130Adc into a capacitive load Z=.000503j ohms(~5F @120Hz). Capacitive load does not charge up as it creates a continuous arc. Im coming up with a 28000uF 63V cap for filter. You agree?
If you have calculated the result using the formula and the parameters provided in the above article, then it should be correct.
Hi.
In your worked example, you write “C = I / 2 x f x Vpp” which is, of course, incorrect. The correct formula is the one you write correctly before it:
C=I/(2 x f x Vpp) where I = current (amps), f = frequency (Hertz) and Vpp is the Peak to Trough voltage difference (volts).
Although I note that you say 1 volt is a useful ripple current. I build guitar effects pedals where 1 mV is the desirable ripple current (a guitar signal ranges from less than 1 mV to 1V).
In my guitar amplifier, with a Vsupply of 400V, a Vpp of 20V after the primary filtering capacitor is acceptable. Further filtering (Resistor-Capacitor type) is used to reduce Vpp at the amplifier. As you’re aware, two smaller capacitors, separated by some resistance is more effective than a single capacitor. (e.g. in the example of the amplifier above, a 47uF capacitor is inferior to a 10uF capacitor followed by a 1kohm resistor and another 10uF capacitor (for a load of 100kohm)).
cheers